Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-2q^2 + 22q - 20}{q^2 - 5q - 50} \div \dfrac{q^2 - q}{q^2 + 8q} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2q^2 + 22q - 20}{q^2 - 5q - 50} \times \dfrac{q^2 + 8q}{q^2 - q} $ First factor out any common factors. $p = \dfrac{-2(q^2 - 11q + 10)}{q^2 - 5q - 50} \times \dfrac{q(q + 8)}{q(q - 1)} $ Then factor the quadratic expressions. $p = \dfrac {-2(q - 10)(q - 1)} {(q - 10)(q + 5)} \times \dfrac {q(q + 8)} {q(q - 1)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { -2(q - 10)(q - 1) \times q(q + 8)} { (q - 10)(q + 5) \times q(q - 1)} $ $p = \dfrac {-2q(q - 10)(q - 1)(q + 8)} {q(q - 10)(q + 5)(q - 1)} $ Notice that $(q - 10)$ and $(q - 1)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-2q\cancel{(q - 10)}(q - 1)(q + 8)} {q\cancel{(q - 10)}(q + 5)(q - 1)} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $p = \dfrac {-2q\cancel{(q - 10)}\cancel{(q - 1)}(q + 8)} {q\cancel{(q - 10)}(q + 5)\cancel{(q - 1)}} $ We are dividing by $q - 1$ , so $q - 1 \neq 0$ Therefore, $q \neq 1$ $p = \dfrac {-2q(q + 8)} {q(q + 5)} $ $ p = \dfrac{-2(q + 8)}{q + 5}; q \neq 10; q \neq 1 $